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An element X has the following isotopic composition `.^(200)X : 90%, .^(199)X: 8.0%, .^(202)X:2%` . The Weighed average atomic mass of naturally occurring element X is closet toA. 199 a.m.u.B. 200 a.m.u.C. 201 a.m.u.D. 202 a.m.u. |
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Answer» Correct Answer - B Average atomic mass, `overset(-)(A)=Sigmaf_(i)A_(i)` (`f_(i)`=fractional abundance, `A_(i)`=atomic mass) Atomic mass `=0.90xx200+0.08xx199+0.02xx202` `=180+15.92+4.04~~200` |
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